This seems like a simple question but I can't find a clear answer in the documentation. I wish to solve for the electric field induced by a conductive particle surrounded by an insulating membrane in an external field. Obviously the high frequency AC cases are the most interesting. I've modeled this using the electric currents module.

My question is this: when I use the frequency domain analysis, does COMSOL solve the complex Laplace equation and complex permittivity values? I want to make sure the the effects of frequency are being properly captured here. Is that the main difference between the electrostatics and electric currents physics?

If not, how might I go about seeing how the polarization of my particle changes with frequency?

Yes, in Frequency Domain, all the variables and material properties can be complex numbers. You can just input a complex value for the permittivity. Keep in mind that, when plotting, only the real part of the plotted quantity is shown; if you want to plot the imaginary part just use the imag() operator.

Note that while Electrostatics (in version 4.3a) allows the Frequency study, the equation solved is still the stationary one. The support for Frequency studies has been added to allow coupling to "slow" physics, such as acoustics or mechanics. To perform a "real" frequency domain study of the electric problem, use Electric Currents.
There is a section about this in the "Theory for the Electrostatics interface" chapter of the AC/DC Module user's guide.

When doing the frequency-domain analysis in the electric currents module, is COMSOL already using complex permittivity values?

In other words, if I specify real permittivity and conductivity for a material, and then do a frequency domain analysis, does COMSOL "automatically" solve using the definition of complex permittivity: epsilon* = epsilon - j*sigma/w, where w is the angular frequency of the field? Or do I have to define

Comsol computes the current directly as the sum of induced current density (from the electrical conductivity sigma) and the displacement current density (from the displacement field D). It does not compute an equivalent complex permittivity since it is not required in the Electric Currents physics.
The permittivity you see in (for example) the plots is the value that you wrote in the Current Conservation feature, it does not change if you specify a certain sigma.

This, of course, applies only to the postprocessing. The actual solution is the same if you specify separately sigma and epsilonr or if you use an equivalent complex permittivity.

i have a question about the electric curruents model. My simulation is 2D axisymmetric in frequency domain.
The geometry is a solenoid which has two turns. As you know, if the current flows through solenoid, the currents in each turn are not the same because of the parasitic capacitance between them. Now, i want to find the current which flows through that capacitance between the two turns (i think that current is displacement current, but i am not sure ). By using "Terminal-terminal type Voltage", i can get a result. But i can't find the current which flows through each turn(ec.Jphi=0). What i want to get is the relationship: ec.Jphi + ec.Jdz = ec.I0_1 Obviously i am worng :( How is the Terminal current ec.I0_1 calculated? what is reacf(V) function? why can i get a induced current (in electric currents model, i think the equation rot(E)= -dB/dt is not used)?

I tried to use "terminal-terminal type Current", i though maybe that can help me. But in electric currents model, i can only get a result by using "Terminal-terminal type Voltage". If i use "terminal-terminal type Current", i get always:

Failed to find a solution
The relative residual(14) is greater than the relative tolerance
Returned solution is not converged.